| Karim Belabas on Mon, 11 Nov 2024 08:51:27 +0100 |
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| Re: How to calculate the conductor of an abelian extension such as Q[x]/(x^3- 19x -19) |
* David Bernier [2024-11-11 05:08]:
> I'm interested in cubic extensions of Q that are abelian, in connection with
> a probable prime test. I have a list of cubic polynomials f_1, ... f_22 and
> I want to find the first f_i such that f_i is irreducible over F_p, where p
> can be assumed prime. For a given f_i, I noticed a periodicity in p of the
> irreducibility character of f_i over F_p (ref. Mathematics Stack Exchange at
> the link: https://math.stackexchange.com/questions/4995484/irreducibility-of-cubic-polynomials-over-finite-fields-f-p
> ). User leoli1 mentioned as relevant the conductor N of the splitting field
> of f_i. I have f_8 = X^3 - 19X - 19 with discriminant 133^2.
(You meant 19^2.)
> How could I calculate the conductor of Q[x]/(f_8) in PARI/gp?
In general (this would work for an abelian extension of an arbitrary
base number field):
Q = bnfinit(y); \\ = field of rational numbers
f8 = x^3 - 19*x - 19; \\ assumed abelian over Q
rnfconductor(Q, f8)[1] \\ or rnfconductor(Q, f8, 2)[1]
the answer is [Mat(19), [0]], which is a pair [f0, foo], where
- foo is the infinite part of the conductor (here trivial),
- f0 is the finite part as an ideal of the maximal order (= the ring Z)
of the base field (here 19 Z)
However, in this particular case of a cyclic cubic number field K, it is
know (cf., e.g., Hasse [1] but it is a 5 lines proof ... assuming class
field theory) that
- foo is always trivial
- f0 is of the form f.Z where f^2 is the discriminant of K.
Hence, f = sqrtint(nfdisc(f8)) gives the answer, in a rather simpler way.
[1] Hasse, H., Arithmetische Theorie der kubischen Zahlkörper auf
klassenkörpertheoretischer Grundlage, Math. Zeitschrift (31), 1930,
pp. 565-582.
Cheers,
K.B.
--
Pr. Karim Belabas, U. Bordeaux, Vice-président en charge du Numérique
Institut de Mathématiques de Bordeaux UMR 5251 - (+33) 05 40 00 29 77
http://www.math.u-bordeaux.fr/~kbelabas/